How do you find the linearization L(x) of f(x) = 3x^2+5x+12 at x=3?

What are the steps for this?

Question

How do you find the linearization L(x) of f(x) = 3x^2+5x+12    at x=3?

What are the steps for this?

myininaya · Answer

We need to find the tangent line of f at x=3

myininaya · Answer

$$f'(x)=6x+15$$
so to find the slope use f'
and then we also know  a point on the line
$$(3,3(3)^2+5(3)+12=27+15+12=54)$$

myininaya · Answer

$$y=mx+b => 54=23(3)+b \text{ solve for b}$$

anonymous · Answer

Wouldn't it be f'(x) = 6x + 17, when you take the derivitive?

myininaya · Answer

oops i put an extra 1 in there lol

myininaya · Answer

$$f'(x)=3(2x)+5=6x+5$$

myininaya · Answer

$$(x^2)'=2x;(5x)'=5;(constant)'=0$$

anonymous · Answer

Ok sounds good, also have another quick question, Where did you get the 23 from...  on
y=mx+b=>54=23(3)+b solve for b

myininaya · Answer

well 23 was what i found when i plug into f'
you know what i got earlier for f' lol

myininaya · Answer

$$f'(3)=6(3)+5=18+5=23$$

myininaya · Answer

f'(3) is the slope of the curve at x=3